Sources separated by 20 m vibrate according to the equation `y_1 = 0.06sin pit` and `y_2 = 0.02 sin pit`. They send out waves along a rod with speed `3m//s` What is the equation of motion of a particle 12 m from the first source and 8 m from the second, `y_1,y_2` are in m?

Explanation

To solve the problem step by step, we will follow the outlined procedure:

### Step 1: Identify the given equations of motion for the two sources
The equations of motion for the two sources are given as:
- \( y_1 = 0.06 \sin(\pi t) \)
- \( y_2 = 0.02 \sin(\pi t) \)

### Step 2: Determine the wave number \( k \)
The wave number \( k \) is calculated using the formula:
\[ k = \frac{\omega}{v} \]
where \( \omega \) is the angular frequency and \( v \) is the speed of the wave. Here, \( \omega = \pi \) (from the sine function) and \( v = 3 \, \text{m/s} \).

Calculating \( k \):
\[ k = \frac{\pi}{3} \]

### Step 3: Write the equations of motion for the two sources at the given distances
We need to find the equations of motion for a particle located 12 m from the first source and 8 m from the second source. The general form of the wave equation is:
\[ y = A \sin(\omega t - kx) \]

For the first source (12 m away):
\[ y_1 = 0.06 \sin(\pi t - k \cdot 12) = 0.06 \sin\left(\pi t - \frac{\pi}{3} \cdot 12\right) = 0.06 \sin\left(\pi t - 4\pi\right) \]
Since \( \sin(x) \) has a periodicity of \( 2\pi \), we can simplify:
\[ y_1 = 0.06 \sin(\pi t) \]

For the second source (8 m away):
\[ y_2 = 0.02 \sin(\pi t - k \cdot 8) = 0.02 \sin\left(\pi t - \frac{\pi}{3} \cdot 8\right) = 0.02 \sin\left(\pi t - \frac{8\pi}{3}\right) \]

### Step 4: Simplify the equations
Now we need to simplify \( y_2 \):
\[ y_2 = 0.02 \sin\left(\pi t - \frac{8\pi}{3}\right) \]
We can express \( \frac{8\pi}{3} \) in terms of \( 2\pi \):
\[ \frac{8\pi}{3} = 2\pi + \frac{2\pi}{3} \quad \text{(which is equivalent to } \frac{2\pi}{3} \text{)} \]
Thus:
\[ y_2 = 0.02 \sin\left(\pi t - \frac{2\pi}{3}\right) \]

### Step 5: Combine the two equations
The total displacement \( y \) at the point is given by:
\[ y = y_1 + y_2 \]
Substituting the expressions we found:
\[ y = 0.06 \sin(\pi t) + 0.02 \sin\left(\pi t - \frac{2\pi}{3}\right) \]

### Step 6: Use the sine addition formula
Using the sine addition formula \( \sin(a - b) = \sin(a)\cos(b) - \cos(a)\sin(b) \):
\[ y_2 = 0.02 \left( \sin(\pi t) \cos\left(\frac{2\pi}{3}\right) - \cos(\pi t) \sin\left(\frac{2\pi}{3}\right) \right) \]
Calculating \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \) and \( \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \):
\[ y_2 = 0.02 \left( \sin(\pi t) \left(-\frac{1}{2}\right) - \cos(\pi t) \left(\frac{\sqrt{3}}{2}\right) \right) \]
\[ y_2 = -0.01 \sin(\pi t) - 0.01\sqrt{3} \cos(\pi t) \]

### Step 7: Combine \( y_1 \) and \( y_2 \)
Now, substituting back into the equation for \( y \):
\[ y = 0.06 \sin(\pi t) - 0.01 \sin(\pi t) - 0.01\sqrt{3} \cos(\pi t) \]
\[ y = (0.06 - 0.01) \sin(\pi t) - 0.01\sqrt{3} \cos(\pi t) \]
\[ y = 0.05 \sin(\pi t) - 0.01\sqrt{3} \cos(\pi t) \]

### Final Equation of Motion
Thus, the final equation of motion for the particle is:
\[ y = 0.05 \sin(\pi t) - 0.01\sqrt{3} \cos(\pi t) \]